What is the theoretical maximum symmetrical fault current that would be available at the secondary terminals of a 3-phase 12,470-480Y/277V, 1000kva, 6.25% impedance transformer?
I am having problems solving this question. Please advise.
What is the theoretical maximum symmetrical fault current that would be available at the secondary terminals of a 3-phase 12,470-480Y/277V, 1000kva, 6.25% impedance transformer?
I am having problems solving this question. Please advise.
Fault Current = Full Load Amps (FLA) / % Impedance (Z)
FLA = VA / L-L Voltage x 1.732
Here is a thread that explains the equations step by step -> http://testguy.net/threads/1341-Easy...-fault-current
Here the information:
12,470-480Y/277V
1000kVA, 6.25% impedance
another way:
For a impedance of 6.25%, your necessary voltage will be 6.25% of 480 Volts = 30 Volts
What that means:
With secondary shorted, if you apply 30 Volts, you will have full transformer current, which is:
(remeber to multiply by 1000 (KVA to VA conversion)
I = 1202.85 A
from here you calculate your transformer Impedance
Z = V/I......Z = 30/1202.85.......Z = 0.025 Ohms.
Your max short-circuit current:
I = V/Z.......I = 480/0.025 = 19200 A
Thanks for the help.
Winding test results on a motor indicate possible degradation when the output waveforms show:
Winding waveform symmetry
Noticeable difference in winding waveforms
Chopped waveforms on grounded windings
Chopped waveforms on phase ungrounded windings
Having trouble getting an answer for this.